E2 reactions are bimolecular and show anti-elimination.

Solution

Chemistry ma’am · 4 steps

Both halves are true: the base and the substrate meet in one step, and H and X leave from opposite faces.

1. One concerted step. In E2 the base pulls off the β-hydrogen while the C–X bond breaks and the C=C bond forms, all in a single step. No intermediate is formed.

   Rate = k [R–X] [Base]

2. Why bimolecular. Two species, the substrate and the base, take part in that slow step. So the rate depends on both concentrations, and the reaction is second order and bimolecular.
3. Why anti. The β-H and the leaving group X must lie in one plane on opposite sides of the C–C bond (anti-periplanar, dihedral angle 180°). Only then do the C–H and C–X orbitals line up to form the new π bond, and this staggered arrangement is also the most stable one.
4. So. E2 is bimolecular and shows anti-elimination. The statement is true.

The idea to keep. E2 means one step with two partners, so rate = k[substrate][base]. Always look for an H that is anti-periplanar to the leaving group; in cyclohexanes both must be axial.

Answer from the NTA final answer key. Solution by PrepWiser.